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Is the ring $A=\mathbb{Z}[T_{\alpha}]$, a ring, such that any ring can be expressed as a subring/quotientring of $A$?

The only rings I am concerned about are commutative rings with 1.

This question occured to me when reading local fields by Serre. In Thm 7 of Chapter II he states that

But if the theorem is proved for one ring $A$, it is also valid for every subring and every quotient ring. As it is true for every polynomial ring $\mathbb{Z}'[T_{\alpha}]=\mathbb{Z}[p^{-1}; T_{\alpha}]$ it holds for $\mathbb{Z}[T_{\alpha}]$ hence for all rings.

implying that every ring can be interpreted as a subring or quotient ring of $\mathbb{Z}[T_{\alpha}]$. How could one formally proof this?

I am aware how to construct the rational, real and complex numbers as well as any extensions from this, but have never gotten in touch with the thought of constructing all and every ring. Or is it just implied by the fact that i can construct $\mathbb{C}$ as an algebraic closed ring and all its extensions?

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    $\begingroup$ Let $R$ be a commutative ring with unit. Then $\mathbb Z\to R$ by $n\mapsto n\cdot 1_R$, and use the universal property of polynomial rings to send $T_a$ to $a\in R$, for all $a\in R$. So $R$ is a quotient ring of some $A$. It's not true that every ring is a subring of some $A$ since not every ring is an integral domain. $\endgroup$
    – user26857
    Oct 8, 2016 at 18:58
  • $\begingroup$ So one constructs a huge polynomial ring, represting each element of a ring as a seperate variable and which gives a surjective homomorphism, which gives the ring as a quotient ring... a lot easier than i thought, thank you! the subring is included to get from $\mathbb{Z}'$ to $\mathbb{Z}$ i think. so it isn't needed for my question $\endgroup$
    – Stefan
    Oct 8, 2016 at 19:13

2 Answers 2

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Every ring $A$ is canonically the quotient of a polynomial ring, namely: $$ \mathbb Z[T_a\vert a\in A] \twoheadrightarrow A: T_a\mapsto a $$

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  • $\begingroup$ I now see that in the comments to the question @user26857 had come up with the same idea before I posted my answer. $\endgroup$ Oct 8, 2016 at 19:23
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It's similar to the statement about any group as the quotient of a free group.

Let $T_{\alpha} = \{\bar{x_0}, \bar{x_1}, \dots \} $ be a set of 'letters' for generators of $\{x_0, x_1, \dots \} \subset A$ and define a map $\mathbb{Z}[T_{\alpha}] \to A$ by $\bar{x} \mapsto x$ for each $\bar{x} \in T_{\alpha}$. This is a surjective homomorphism, so...

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    $\begingroup$ Your notation is very weird. $\endgroup$ Oct 8, 2016 at 19:04
  • $\begingroup$ Ya $\bar{x}$ or something would have been better but I think it's still clear. $\endgroup$
    – basket
    Oct 8, 2016 at 19:07
  • $\begingroup$ OK I'll fix it haha $\endgroup$
    – basket
    Oct 8, 2016 at 19:08

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